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definition of observer and time measured by different observers in general relativity
Why do clocks measure arc-length?Are there more distinctive names of “null curves” with certain additional properties (absence of “chord curves”)?An argument that massive particles don't redshift?How to determine the three-velocity measured by a single observer?Global symmetries of spacetime and general covarianceHow a reference frame relates to observers and charts?When the $x^0$ coordinate represents time in GR?How to make sense of this definition of a reference frame?How one uses the definition of observers in General Relativity?How do we measure distances in the FLRW metric?
$begingroup$
An observer in general relativity is defined as a future directed timelike worldline
beginalign*
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
endalign*
together with an orthonormal basis $e_a(lambda) in T_gamma(lambda)M$ where $e_0(lambda)= v_gamma, gamma(lambda)$ and
beginalign
g_gamma(lambda)(e_a(lambda),e_b(lambda))=eta_ab~. qquad (1)
endalign
Here, $v_gamma, gamma(lambda)$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
beginalign
tau_gamma = int_lambda_0^lambda_1 dlambda sqrtg_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g_gamma(lambda)(e_0(lambda),e_0(lambda))=1 qquad (2)
endalign
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
beginalign*
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
endalign*
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
beginalign
tau_delta = int_lambda_0^lambda_1 dlambda sqrtg_delta(lambda)(v_delta, delta(lambda),v_delta, delta(lambda))~.
endalign
From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.
However, I know that my conclusion is wrong. Can you point out where I went astray?
general-relativity observers
asked 10 hours ago
damaihatidamaihati
683
$endgroup$
add a comment |
$begingroup$
An observer in general relativity is defined as a future directed timelike worldline
beginalign*
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
endalign*
together with an orthonormal basis $e_a(lambda) in T_gamma(lambda)M$ where $e_0(lambda)= v_gamma, gamma(lambda)$ and
beginalign
g_gamma(lambda)(e_a(lambda),e_b(lambda))=eta_ab~. qquad (1)
endalign
Here, $v_gamma, gamma(lambda)$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
beginalign
tau_gamma = int_lambda_0^lambda_1 dlambda sqrtg_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g_gamma(lambda)(e_0(lambda),e_0(lambda))=1 qquad (2)
endalign
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
beginalign*
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
endalign*
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
beginalign
tau_delta = int_lambda_0^lambda_1 dlambda sqrtg_delta(lambda)(v_delta, delta(lambda),v_delta, delta(lambda))~.
endalign
From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.
However, I know that my conclusion is wrong. Can you point out where I went astray?
general-relativity observers
asked 10 hours ago
damaihatidamaihati
683
$endgroup$
add a comment |
$begingroup$
An observer in general relativity is defined as a future directed timelike worldline
beginalign*
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
endalign*
together with an orthonormal basis $e_a(lambda) in T_gamma(lambda)M$ where $e_0(lambda)= v_gamma, gamma(lambda)$ and
beginalign
g_gamma(lambda)(e_a(lambda),e_b(lambda))=eta_ab~. qquad (1)
endalign
Here, $v_gamma, gamma(lambda)$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
beginalign
tau_gamma = int_lambda_0^lambda_1 dlambda sqrtg_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g_gamma(lambda)(e_0(lambda),e_0(lambda))=1 qquad (2)
endalign
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
beginalign*
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
endalign*
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
beginalign
tau_delta = int_lambda_0^lambda_1 dlambda sqrtg_delta(lambda)(v_delta, delta(lambda),v_delta, delta(lambda))~.
endalign
From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.
However, I know that my conclusion is wrong. Can you point out where I went astray?
general-relativity observers
asked 10 hours ago
damaihatidamaihati
683
$endgroup$
An observer in general relativity is defined as a future directed timelike worldline
beginalign*
gamma:I subset mathbb R &to M \
lambda &mapsto gamma(lambda)
endalign*
together with an orthonormal basis $e_a(lambda) in T_gamma(lambda)M$ where $e_0(lambda)= v_gamma, gamma(lambda)$ and
beginalign
g_gamma(lambda)(e_a(lambda),e_b(lambda))=eta_ab~. qquad (1)
endalign
Here, $v_gamma, gamma(lambda)$ is the velocity of the worldline $gamma$ at the point $gamma(lambda)in M$ and $g$ is the metric tensor field on $M$. The time measured by the clock carried by this observer between events $lambda_0, lambda_1$ is defined as
beginalign
tau_gamma = int_lambda_0^lambda_1 dlambda sqrtg_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda))~.
endalign
However,
beginalign
g_gamma(lambda)(v_gamma, gamma(lambda),v_gamma, gamma(lambda)) = g_gamma(lambda)(e_0(lambda),e_0(lambda))=1 qquad (2)
endalign
which follows from the requirement of eq.(1). We are using signature $(+,-,-,-)$.
This is all standard definition. Suppose, we have another observer $delta$:
beginalign*
delta:I subset mathbb R &to M \
lambda &mapsto delta(lambda)
endalign*
and the time measured by his clock between the same two events $lambda_0, lambda_1$ is
beginalign
tau_delta = int_lambda_0^lambda_1 dlambda sqrtg_delta(lambda)(v_delta, delta(lambda),v_delta, delta(lambda))~.
endalign
From equations (1) and (2), we get $tau_gamma = tau_delta$ and this will be true for all observers measuring time between $lambda_0, lambda_1$.
However, I know that my conclusion is wrong. Can you point out where I went astray?
general-relativity observers
general-relativity observers
asked 10 hours ago
damaihatidamaihati
683
asked 10 hours ago
damaihatidamaihati
683
asked 10 hours ago
damaihatidamaihati
683
asked 10 hours ago
damaihatidamaihati
683
asked 10 hours ago
damaihatidamaihati
683
683
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Your conclusion is correct, because what you are doing by saying that $g(v_gamma,gamma(lambda),v_gamma,gamma(lambda)) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
answered 10 hours ago
VoidVoid
10.7k1757
$endgroup$
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
4 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
Your conclusion is correct, because what you are doing by saying that $g(v_gamma,gamma(lambda),v_gamma,gamma(lambda)) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
answered 10 hours ago
VoidVoid
10.7k1757
$endgroup$
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
4 hours ago
add a comment |
$begingroup$
Your conclusion is correct, because what you are doing by saying that $g(v_gamma,gamma(lambda),v_gamma,gamma(lambda)) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
answered 10 hours ago
VoidVoid
10.7k1757
$endgroup$
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
4 hours ago
add a comment |
$begingroup$
Your conclusion is correct, because what you are doing by saying that $g(v_gamma,gamma(lambda),v_gamma,gamma(lambda)) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
answered 10 hours ago
VoidVoid
10.7k1757
$endgroup$
Your conclusion is correct, because what you are doing by saying that $g(v_gamma,gamma(lambda),v_gamma,gamma(lambda)) = 1$ is that the parameter $lambda$ is exactly equal to proper time. You can have different parametrizations $tildelambda$ of the curve $gamma$ that have $g(v_gamma,gamma(tildelambda),v_gamma,gamma(tildelambda)) neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
answered 10 hours ago
VoidVoid
10.7k1757
answered 10 hours ago
VoidVoid
10.7k1757
answered 10 hours ago
VoidVoid
10.7k1757
answered 10 hours ago
VoidVoid
10.7k1757
10.7k1757
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
4 hours ago
add a comment |
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
4 hours ago
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
4 hours ago
$begingroup$
I think maybe the answer could be phrased differently: there is a mistake in the OP's math, which is that the limits of integration have no particular reason to be the same. I think the point is made more clearly, but that may just be me.
$endgroup$
– Javier
4 hours ago
add a comment |
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